Question

The compound X(C5H8) reacts with ammonical AgNO3 to give a white precipitate and on oxidation with hot alkaline KMnO4 gives the acid (CH3)2CHCOOH. Therefore X is

Answer 1

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Answer 2

Answer:

Reaction with hot$\mathrm{KMnO}_{4}$ is:

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}+\mathrm{CO}_{2}$

Explanation:

• Terminal alkynes contain acidic hydrogen. These terminal alkynes react with ammoniacal solutions of silver nitrate to give white precipitate of silver alkynide.
• Since compound ' $X^{\prime}$on reacting with ammoniacal $\mathrm{AgNO}_{3}$ gives white precipitate. The given compound will have terminal alkyne part.
• From the given options we can see that compound of option (c) has terminal alkyne part.
• $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \mathrm{NO}_{3}^{-} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Ag}^{+} \downarrow+\mathrm{NH}_{4}^{+} \mathrm{NO}_{3}^{-}+$

• When this compound (c) is treated with excesss $\mathrm{KMnO}_{4}$, the cleavage will occur at alkyne part and they get oxidise to form carboxylic acid. Thus, it gives $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{COOH}$and $\mathrm{CH}_{3} \mathrm{COOH}$
• $\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}-\mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{KMnO}_{4}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}-\mathrm{COOH}+\mathrm{HCOOH}$